測試數據非常強大，一兩個錯誤就猛吃 WA
利用公式解可以解出這道題目，只要利用那兩條等式換來換去就可以得到 ay4+by2+c=0 這個等式
再套公式即可。
//====================================================================||
// Name : A Day in Math-land.cpp ||
// Date : 2013/7/18 下午11:20:06 ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 10001
#define PII pair<int,int\>
#define PB push\_back
#define oo (1LL<<62)
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
Int p,q;
inline Int sqr(Int x){
return x\*x;
}
inline bool isqrt(Int x,Int &ans){
if(x<0)return false;
ans=sqrt(x);
return ans\*ans==x;
}
void solve(){
Int d=sqr(p)+6LL\*p\*q+sqr(q);
Int y,x,tmp,t,ansx=oo,ansy;
bool is\=isqrt(d,t);
if(!is){
printf("Impossible.\\n");
return ;
}
// debug("%lld\\n",t);
if((3LL\*p+q+t)%4==0){
tmp=(3LL\*p+q+t)>>2LL;
is\=isqrt(tmp,y);
if(is){
bool fg=false;
if(y&&p%y==0){
x=p/y-y;
fg=true;
}else if(y==0){
fg=isqrt(q,x);
}
if(x>=y&&x<ansx&&fg){
ansx=x;
ansy=y;
}
fg=false;
y=-y;
if(y&&p%y==0){
x=p/y-y;
fg=true;
}else if(y==0){
fg=isqrt(q,x);
}
if(x>=y&&x<ansx&&fg){
ansx=x;
ansy=y;
}
}
}
if((3LL\*p+q-t)%4==0){
tmp=(3LL\*p+q-t)>>2LL;
is\=isqrt(tmp,y);
if(is){
bool fg=false;
if(y&&p%y==0){
x=p/y-y;
fg=true;
}else if(y==0){
fg=isqrt(q,x);
}
if(x>=y&&x<ansx&&fg){
ansx=x;
ansy=y;
}
fg=false;
y=-y;
if(y&&p%y==0){
x=p/y-y;
fg=true;
}else if(y==0){
fg=isqrt(q,x);
}
if(x>=y&&x<ansx&&fg){
ansx=x;
ansy=y;
}
}
}
if(ansx==oo)printf("Impossible.\\n");
else printf("%lld %lld\\n",ansx,ansy);
}
int main() {
ios\_base::sync\_with\_stdio(0);
int test;
scanf("%d",&test);
int ff=0;
while(test--){
scanf("%lld %lld",&p,&q);
printf("Case %d:\\n",++ff);
solve();
}
}